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prove that
modulus of sin(nx)<= n* modulus of sin(x)
for n being a natural number.
Sketch the graphs of both the fucntions ,
n*mod(Sin(x)) is a curve with amplitude n and period 2pi
mod(Sin(nx)) is a curve with amplitude 1 and period 2pi/n
Hi Aritra,
We can plot a graph and see that the above inequality is true.
But to actually plot the graph we need to do some mathematical analysis.
We can prove the direct inequality mathematically.
Consider |sin(nx)/sinx| = |sin[(n-1)x+x]/sinx| = |cosx + sin(n-1)xcosx/sinx|
Now write sin(n-1)x as sin[(n-2)x+x] and expand again.
We will have |cos(n-1)x + cosx*cos(n-2)x + cos^2 x*sin(n-2)x/sinx|....
We keep on doing this till we have the summation of n terms within the modulus.
Clearly each term is a trignometric term in the form of either sin/cos, and hence each term is less than 1.
Hence the summation of the n terms will be less than n.
So clearly |sin(nx)/sinx| ≤ |n| = n.
And hence |sin(nx)| ≤ n|sinx|.
Hope this helps.
All the best.
Regards,
Ashwin (IIT MadraS).
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